---
title: "Problem Set 2"
notoc: true
---

# Problem Set 2


:::{.exercise title="Gathmann 2.17"}
Find the irreducible components of 
\[  
X = V(x - yz, xz - y^2) \subset \AA^3/\CC
.\]
:::

:::{.solution}
Since $x=yz$ for all points in $X$, we have
\[  
X &= V(x-yz, yz^2 - y^2) \\
&= V\qty{x-yz, y(z^2 - y) } \\
&= V(x-yz, y) \union V(x-yz, z^2-y) \\
&\da X_1 \union X_2
.\]

:::{.claim}
These two subvarieties are irreducible.
:::

It suffices to show that the $A(X_i)$ are integral domains.
We have
\[  
A(X_1) \da \CC[x,y,z] / \gens{x-yz, y} \cong \CC[y,z]/\gens{y} \cong \CC[z]
,\]
which is an integral domain since $\CC$ is a field and thus an integral domain, and
\[  
A(X_2) \da \CC[x,y,z]/\gens{x-yz, z^2 - y} \cong \CC[y,z]/\gens{z^2-y} \cong \CC[y]
,\]
which is an integral domain for the same reason.




:::



:::{.exercise title="Gathmann 2.18"}
Let $X\subset \AA^n$ be an arbitrary subset and show that 
\[  
V(I(X)) = \bar{X}
.\]

:::

:::{.solution}
\hfill

$\bar X \subseteq V(I(X))$:
\
We have $X\subseteq V(I(X))$ and since $V(J)$ is closed in the Zariski topology for any ideal $J \normal \kx{n}$ by definition, $V(I(X))$ is closed.
Thus 
\[  
X\subseteq V(I(X)) \text{ and } V(I(X))\text{ closed } \implies \bar X \subseteq V(I(X))
,\]
since $\bar X$ is the intersection of all closed sets containing $X$.
\

$V(I(X)) \subseteq \bar X$:
\
Noting that $V(\wait), I(\wait)$ are individually order-reversing, we find that $V(I(\wait))$ is order-*preserving* and thus
\[  
X\subseteq \bar X \implies V(I(X)) \subseteq V(I(\bar X)) = \bar X
,\]
where in the last equality we've used part (i) of the Nullstellensatz: if $X$ is an affine variety, then $V(I(X)) = X$.
This applies here because $\bar X$ is always closed, and the closed sets in the Zariski topology are precisely the affine varieties.



:::


:::{.exercise title="Gathmann 2.21"}
Let $\ts{U_i}_{i\in I} \covers X$ be an open cover of a topological space with $U_i \intersect U_j \neq \emptyset$ for every $i, j$.

a. Show that if $U_i$ is connected for every $i$ then $X$ is connected.

b. Show that if $U_i$ is irreducible for every $i$ then $X$ is irreducible.
:::

:::{.solution title="a"}
Suppose toward a contradiction that $X = X_1 \disjoint X_2$ with $X_i$ proper, disjoint, and open.
Since $\ts{U_i} \covers X$, for each $j\in I$ this would force one of $U_j \subseteq X_1$ or $U_j \subseteq X_2$, since otherwise $U_j \intersect X_1 \intersect X_2$ would be nonempty.
\
So without loss of generality (relabeling if necessary), assume $U_j \in X_1$ for some fixed $j$.
But then for every $i\neq j$, we have $U_i \intersect U_j$ nonempty by assumption, and so in fact $U_i \subseteq X_1$ for every $i\in I$.
But then $\union_{i\in I}U_i \subseteq X_1$, and since $\ts{U_i}$ was a cover, this forces $X\subseteq X_1$ and thus $X_2 = \emptyset$.
:::


:::{.solution title="b"}
:::{.claim}
$X$ is irreducible $\iff$ any two open subsets intersect.
:::
This follows because otherwise, if $U, V \subset X$ are open and disjoint then $X\sm U,\, X\sm V$ are proper and closed.
But then we can write $X = \qty{X\sm U} \disjoint \qty{X\sm V}$ as a union of proper closed subsets, forcing $X$ to not be irreducible.
\

So it suffices to show that if $U, V\subset X$ then $U\intersect V$ is nonempty.
Since $\ts{U_i} \covers X$, we can find a pair $i, j$ such that there is at least one point in $U\intersect U_i$ and one point in $V \intersect U_j$.
\

But by assumption $U_i\intersect U_j$ is nonempty, so both $U\intersect U_i$ and $U_j \intersect U_i$ are open nonempty subsets of $U_i$.
Since $U_i$ was assumed irreducible, they must intersect, so there exists a point
\[  
x_0 \in \qty{U\intersect U_i} \intersect \qty{U_j \intersect U_i} = U\intersect \qty{U_i \intersect U_j} \da \tilde U
.\]

We can now similarly note that $\tilde U \intersect V$ and $U_j \intersect V$ are nonempty open subsets of $V$, and thus intersect.
So there is a point
\[  
\tilde x_0 \in \qty{\tilde U \intersect V} \intersect \qty{U_j \intersect V} = \tilde U\intersect V = U\intersect V \intersect \qty{U_i \intersect U_j}
,\]
and in particular $\tilde x_0 \in U\intersect V$ as desired.

:::


:::{.exercise title="Gathmann 2.22"}
Let $f:X\to Y$ be a continuous map of topological spaces.

a. Show that if $X$ is connected then $f(X)$ is connected.

b. Show that if $X$ is irreducible then $f(X)$ is irreducible.

:::

:::{.solution title="a"}
Toward a contradiction, if $f(X) = Y_1 \disjoint Y_2$ with $Y_1, Y_2$ nonempty and open in $Y$, then 
$$f\inv(f(X)) \subseteq X$$ 
on one hand, and 
$$f^{-1}(f(X)) = f^{-1}(Y_1) \disjoint f^{-1}(Y_2)$$
on the other.
If $f$ is continuous, the preimages $f^{-1}(Y_i)$ are open (and nonempty), so $X$ contains a disconnected subset.
However, every subset of a connected set must be connected, so this contradicts the connectedness of $X$.
:::

:::{.solution title="b"}
Suppose $f(X) = Y_1 \union Y_2$ with $Y_i$ proper closed subsets of $Y$.
Then $f^{-1}(Y_1) \union f^{-1}(Y^2) = (f^{-1} \circ f)(X) \subseteq X$ are closed in $X$, since $f$ is continuous.
Since $X$ is irreducible, without loss of generality (by relabeling), this forces $X_1 = \emptyset$.
But then $f(X_1) = \emptyset$, forcing $f(X) = Y_2$.
:::



:::{.definition title="Ideal Quotient"}
For two ideals $J_1, J_2\normal R$, the *ideal quotient* is defined by
\[  
J_1 : J_2 \da \ts{f\in R \st fJ_2 \subset J_1}
.\]

:::


:::{.exercise title="Gathmann 2.23"}
Let $X$ be an affine variety.

a. Show that if $Y_1, Y_2 \subset X$ are subvarieties then 
\[  
I(\bar{Y_1\sm Y_2}) = I(Y_1): I(Y_2)
.\]

b. If $J_1, J_2 \normal A(X)$ are radical, then
\[  
\bar{V(J_1) \sm V(J_2)} = V(J_1: J_2)
.\]

:::

:::{.solution}
?
:::


:::{.exercise title="Gathmann 2.24"}
Let $X \subset \AA^n,\, Y\subset \AA^m$ be irreducible affine varieties, and show that $X\cross Y\subset \AA^{n+m}$ is irreducible.
:::

:::{.solution}
That $X\cross Y$ is again an affine variety follows from writing $X=V(I),\, Y=V(J)$, then $X\cross Y = V(I+J)$ where $I+J\normal k[x_1, \cdots, x_n, y_1, \cdots, y_m]$.
So let 
$$X\cross Y = U \union V$$ 
with $U, V$ proper and closed, and let $\pi_X, \pi_Y$ be the projections onto the factors.

:::{.claim}
For each $x\in X$, $\pi^{-1}(x) \cong Y$ is contained in only one of $U$ or $V$.
:::

Note that if this is true, we can write $X = G_U \union G_V$ where 
\[  
G_U\da\ts{x\in X \st \pi_X^{-1}(x) \subseteq U}
\]
are the points for which the entire fiber lies in $U$, and similarly $G_V$ are those for which the fiber lies in $V$.
If we can then show that $G_U, G_V$ are closed, by irreducibility of $X$ this will force (wlog) $G_V = \emptyset$ and $X = G_U$.
But then 
\[  
\pi_X^{-1}(X) = X\cross Y \text{ and }\pi_X^{-1}(G_U) = U  \implies X\cross Y = U
.\]
which shows that $X\cross Y$ is irreducible.

:::{.proof title="Every fiber is contained in one irreducible component"}
For any fixed $x$, we can write 
\[  
\pi_X^{-1}(x) = \qty{\pi_X\inv(x) \intersect U } \union \qty{\pi_X^{-1}(x) \intersect V}
.\]

Since points are closed in the Zariski topology and $\pi_X$ is continuous, each $\pi_X^{-1}(x)$ is closed.
and thus $\pi_X^{-1}(x)\intersect U$ is closed (and similarly for $V$).
Noting that $\pi_X^{-1}(x) \cong \ts{x}\cross Y \cong Y$, where we've assumed $Y$ to be irreducible, we can conclude wlog that $\pi_X^{-1}(x) \intersect V = \emptyset$.
:::

:::{.proof title="$G_U, G_V$ are closed"}
Wlog consider $G_U \subseteq X$.
Fixing any point $y_0 \in Y$, we have 
$$X\cong X_{y_0} \da X\cross \ts{y_0} \subseteq X\cross Y,$$
so we can identify $G_U \subset X$ with $G_U\subset X_{y_0}$ inside a $Y\dash$fiber the product.
But then 
$$G_U = X_{y_0} \intersect U \subseteq X\cross Y,$$ 
where $U$ is closed in $X\cross Y$ and thus closed in $X_{y_0}$, and $X_{y_0}$ is trivially closed in itself.
This exhibits $G_U$ as the intersection of two sets that are closed in $X_{y_0} \cong X$.
:::


:::